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ECE 305: Spring 2018
Semiconductor Equations
Professor Peter BermelElectrical and Computer Engineering
Purdue University, West Lafayette, IN [email protected]
Pierret, Semiconductor Device Fundamentals (SDF)
Chapter 3 (pp. 122-138)
2/8/18 Bermel ECE 305 S18 1
Bermel ECE 305 S18
outline
1. Semiconductor Equation Overview
2. MCDE Examples
3. Solving Poisson’s Equation
2/8/18
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the semiconductor equations
Five equations in five
unknowns:
r = q p - n + N
D
+ - NA
-( )
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( )J x
( )J x dx
Ng
n
Nr
, 𝐽𝑝 𝐽𝑛
Semiconductor equations: 2 key cases
Diffusion problems (ℇ = 0): MCDE
Drift problems (ℇ ≠ 0): Drift current equations
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¶Dp
¶t= Dp
d2Dp
dx2-
Dp
t p
+ GL
Bermel ECE 305 S18
outline
1. Semiconductor Equation Overview
2. MCDE Examples
3. Solving Poisson’s Equation
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Example #1: Solution
x
Dn x( )
Dn x( ) = GLt n
x = 0
Steady-state, uniform generation, no spatial variation
Dn = GLt n =1020 ´10-6 =1014
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x = L = 200 mm
6
Example #2 Summary
Bermel ECE 305 S18
1
JP P P
pr g
t q
D AD q p n N N
P P Pqp qD p J E
1N N N
nr g
t q
J
N N Nqn qD n J E
1
time
Dn
time
Analytical solutions
ntt
entxn
DD 0,
( , ) 1 nt
nn x t G e
D
Dn
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Example #3: One-sided Carrier Diffusion
Bermel ECE 305 S18
1 nN N
n dJr g
t q dx
N N N
dnqn E qD
dx J
2
20 N
d nD
dx
Steady state, no generation/recombination
1
0,' D txn
a 0x’
Metal contact
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Example 3B: One-sided carrier diffusion
Long sample (steady state)
Steady-state, sample long compared to the diffusion length.
i.e., a short diffusion length
fixedDp x = 0( )
1) Simplify the MCDE
2) Solve the MCDE
3) Deduce Fp from Δp
¶Dp
¶t= Dp
d2Dp
dx2-
Dp
t p
+ GL
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Example 3B: One-sided carrier diffusion
Long sample (steady state)
Steady-state, sample long compared to the diffusion
length. No generation. fixedDn x = 0( ) =1012 cm-3
Step 4B ) Key words: steady-state, without light, long device
ii) If I write for MCDE, would I be right?¶Dp
¶t= Dp
d2Dp
dx2-
Dp
t p
+ GL
iii) Which approximate equation should I choose:
Step 3) What type of problems are we talking about?
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Example 3B: One-sided carrier diffusion
Long sample (steady state)
x
Dn x( )
Dn x ®¥( ) = 0
Dn 0( )
Dn x( ) = Dn 0( )e-x/Ln
x = L = 200 mmx = 0
Ln = Dnt n << L
Steady-state, sample long compared to the diffusion length.
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Example 3B: One-sided carrier diffusion
Long sample (steady state)
Steady-state, sample is 5 micron long. No generation.
Which approximate equation should I choose:
Dn x = 0( ) =1012 cm-3
Dn x = 5 mm( ) = 0
What are my boundary conditions?
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Example 3B: One-sided carrier diffusion
Long sample (steady state)
Dn x = 0( ) =1012 cm-3
0Dn x LD
Dn x = 0( ) =1012 cm-3
Dn x( ) = Ax + B
Dn x( ) = Dn 0( ) 1-x
L
æ
èçö
ø÷
0n x LD
2/8/18 Bermel ECE 305 S1814
Example 3C: One-sided carrier diffusion
Intermediate sample (steady state)
Steady-state, sample is 30 micrometers long. No generation.
fixedDn x = 0( ) =1012 cm-3
1) Simplify the MCDE
2) Solve the MCDE
3) Deduce Fp from Δp
¶Dn
¶t= Dp
d2Dn
dx2-
D
t n
+ GL
0 = Dp
d2Dn
dx2-
D
t n
+ 0
d2Dn
dx2-
Dn
Ln
= 0 Ln º Dnt n
Dn x = 30 mm( ) = 0
Ln = 28 mm L= 30 mm
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Example 3C: One-sided carrier diffusion
Intermediate sample (steady state)
Steady-state, sample is 30 micrometers long. No generation.
fixedDn x = 0( ) =1012 cm-3
1) Simplify the MCDE
2) Solve the MCDE
3) Deduce Fp from Δp
d2Dn
dx2-
Dn
Ln
= 0
Dn x = 30 mm( ) = 0Dn x( ) = Ae-x/Ln + Be+x/Ln
Dn 0( ) = A + B =1012
Dn L( ) = Ae-L/Ln + Be+L/Ln = 0
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Example 3C: One-sided carrier diffusion
Intermediate sample (steady state)
x
Dn x( )
Dn x = L( ) = 0
Dn 0( )
x = Lx = 0
Dn x( )= Dn 0( )sinh L - x( ) / Lnéë ùû
sinh L / Ln( )
Steady-state, sample neither long nor short compared to the
diffusion length.
Ln = Dnt n » L
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Example #3 summary: One-sided carrier diffusion
Length scale Solution type
Decaying exponentials
Linear
Hyperbolic functions
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Homework #5: Fall 2017
(cf. this week’s homework!)
1. Assume that a n-type region of crystalline silicon with 𝜇𝑝 = 450 cm2/V ∙ s and lifetime 𝜏𝑝 = 10 ms is uniformly
illuminated by a photon flux 𝐺𝐿 = 1020 /cm3∙s, reaches a steady state, which is then switched off at 𝑡 = 0.
a. Write down the simplest form of the minority carrier diffusion equation that accurately describes its behavior. Briefly justify your answer.
b. Sketch the time-dependent decay of the carrier concentration.
c. What is the value of the carrier concentration at 𝑡 = 20 ms?
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Homework #5: Fall 2017
a. Write down the simplest form of the minority carrier diffusion equation that accurately describes its behavior. Briefly justify your answer.
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Homework #5: Fall 2017a.
b. Sketch the time-dependent decay of the carrier concentration.
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Homework #5: Fall 2017a.
b.
c. What is the value of the carrier concentration at 𝑡 = 20 ms?
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Homework #5: Fall 2017
2. Assume that a uniform p-type region of gallium arsenide (𝑇 = 300 K) has length 𝐿 = 1 m, lifetime 𝜏𝑛 = 5 ns and 𝜇𝑛 = 8500 cm2/V ∙ s. Assume that all carriers are extracted at 𝑥 = 𝐿 (such that ∆𝑛 = 0), while ∆𝑛 = 1013 /cm3 at 𝑥 = 0. Assume that 𝑝𝑜 ≫ ∆𝑛 everywhere. Now consider the system after reaching a steady state.
a. Write down the simplest form of the minority carrier diffusion equation that accurately describes its behavior. Briefly justify your answer.
b. Solve for and sketch the minority carrier concentration ∆𝑛 as a function of position between 𝑥 = 0 and 𝑥 = 𝐿.
c. How would this problem change if the minority carrier lifetime 𝜏𝑛 were reduced to 1 ns? Justify your answer quantitatively.
2/8/18 Bermel ECE 305 S1826
Homework #5: Fall 2017
a. Write down the simplest form of the minority carrier diffusion equation that accurately describes its behavior. Briefly justify your answer.
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Homework #5: Fall 2017a.
b. Solve for and sketch the minority carrier concentration ∆𝑛 as a function of position between 𝑥 = 0 and 𝑥 = 𝐿.
2/8/18 Bermel ECE 305 S1829
Homework #5: Fall 2017a.
b.
c. How would this problem change if the minority carrier lifetime 𝜏𝑛 were reduced to 1 ns? Justify your answer quantitatively.
2/8/18 Bermel ECE 305 S1831
Bermel ECE 305 S18
outline
1. Semiconductor Equation Overview
2. MCDE Examples
3. Solving Poisson’s Equation
2/8/18
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✓
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“the semiconductor equations”
Three equations in three
unknowns:
In steady state equilibrium, we only need to solve the
Poisson equation
How do we calculate rho(x), E(x), and V(x)?
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equilibrium energy band diagrams solve
Poisson’s equation – and more
EFEF
1) Begin with EF
2) Draw the E-bands where you know the carrier density
3) Electrostatic potential by flipping E-band upside down.
4) E-field from slope
5) n(x), p(x) from the E-band diagram
6) rho(x) from n(x) and p(x)
7) diffusion current from (5) or from (6)
EC x( ) = EC-ref - qV x( )
E x( ) =1
qdEC x( ) dx
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energy band diagram
36
EF
EC
EV
x
E
Ei
x = xpx = 0x = -xn
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“read” the e-band diagram
1) Electrostatic potential vs. position
2) Electric field vs. position
3) Electron and hole densities vs. position
4) Space-charge density vs. position
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NP junction (equilibrium)
N P
xp-xn 0
“transition (depletion) region”
Bermel ECE 305 S18
Vbi =kBT
qln
NDNA
ni
2
æ
èçö
ø÷
V = 0V = Vbi
1) What is the width of the depletion region?
2) What is the maximum electric field?
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the Poisson equation
d
dxeSE( ) = r x( )
dE
dx=
r x( )eS
=r x( )KSe0
dE
dx=
r x( )KSe0
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the “depletion approximation”
dE
dx=
r x( )KSe0
r
x
N P
-xn
r = +qND
xp
r = -qNA
qNDxn = qNAxp
NDxn = NAxp
2/8/18 Bermel ECE 305 S18 40
the electric field
dE
dx=
r x( )KSe0
r
xN P
xp
-xn
r = -qNA
dE
dx=
qND
KSe0
dE
dx= -
qNA
KSe0
E x( ) > 0
2/8/18 Bermel ECE 305 S18 41
the electrostatic potential
dE
dx=
r x( )KSe0
xN P
xp-xn
E
E x( ) = -dV
dx
W = xn + xP
E 0( ) =qND
KSe0
xn
NDxn = NAxPxn =NA
NA + ND
W
dE
dx=
qND
KSe0
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the electrostatic potential
dE
dx=
r x( )KSe0
N P
E x( ) = -dV
dx
Vbi = E x( )-xn
xp
ò dx
Vbi =1
2E 0( )W
E 0( ) =qND
KSe0
NA
NA + ND
W
W =2KSe0
q
NA + ND
ND NA
æ
èçö
ø÷Vbi
é
ëê
ù
ûú
1/2
E 0( ) =2qVbi
Kse0
ND NA
NA + ND
æ
èçö
ø÷
é
ëê
ù
ûú
1/2
2/8/18 Bermel ECE 305 S18
xxp-xn
E
ℰ(0)
43
calculating 𝑉(𝑥) from ℰ(𝑥)
xP
xp-xn
E
V x( ) = - E x( )x
xp
ò dx
V x( )
V = 0
V = Vbi
See Pierret, SDF, pp. 212-213
2/8/18 Bermel ECE 305 S18
Vbi = E x( )-xn
xp
ò dx
Vbi =1
2E 0( )W
ℰ(0)
𝑊
E x( ) = -dV
dx
44
summary
xN P
xp-xn
E
W = xn + xP
NDxn = NAxP
xn =NA
NA + ND
W
xp =ND
NA + ND
W
W =2KSe0
q
NA + ND
ND NA
æ
èçö
ø÷Vbi
é
ëê
ù
ûú
1/2
E 0( ) =2qVbi
Kse0
ND NA
NA + ND
æ
èçö
ø÷
é
ëê
ù
ûú
1/2
Vbi =kBT
qln
ND NA
ni
2
æ
èçö
ø÷
E 0( ) =2Vbi
W
2/8/18 Bermel ECE 305 S18 45
example
N P
ND = 1015
depletion region
xp-xn 0
E 0( )
W
NA = 1015
Vbi =kBT
qln
NDNA
ni
2
æ
èçö
ø÷= 0.6 V
W =1.25 mm xn = xp = 0.625 mm
E 0( ) = 9.6 ´103 V/cm
2/8/18 Bermel ECE 305 S18 46
“one-sided junction”
N P
ND =1018 cm-3
0
E 0( )
W
NA = 1015
x
xn , xp
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numbers: one-sided junction
W = 1.010 ´10-4 cm 1.25 ´10-4( )
E 0( ) =2Vbi
W= 1.5 ´104 V/cm (0.96)
N
ND = 1018
0
Vbi = 0.78 V (0.60)
xp-xn
W NA = 1015
xn = 0.001´10-4 cm
xp = 1.009 ´10-4 cm
P
2/8/18 Bermel ECE 305 S18 48
one-sided junction
ND >> NA
xN P
xp
-xn
dE
dx=
qND
KSe0
dE
dx= -
qNA
KSe0
E x( )
xp >> xn
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one-sided junction
W =2KSe0
q
NA + ND
NDNA
æ
èçö
ø÷Vbi
é
ëê
ù
ûú
1/2
®2KSe0
qNA
Vbi
é
ëê
ù
ûú
1/2
E 0( ) =2Vbi
W
xn =NA
NA + ND
W » 0
xp =ND
NA + ND
W » W
N P
ND = 1018
depletion region
xp-xn0
E x( )
Vbi »kBT
qln
ND NA
ni
2
æ
èçö
ø÷
W NA = 1015
2/8/18 Bermel ECE 305 S1850
example: one-sided junction
W =2KSe0
qNA
Vbi
é
ëê
ù
ûú
1/2
E 0( ) =2Vbi
W
xn » 0
xp » W
N P
ND = 1018
xp-xn 0
E 0( )
Vbi »kBT
qln
ND NA
ni
2
æ
èçö
ø÷
W NA = 1016
2/8/18 Bermel ECE 305 S18 51
conclusions We will solve many problems in this class using the
semiconductor equations: In regions of zero field, we can use the minority carrier
diffusion equation to understand the mechanics of carrier transport in electronic devices. Review the problem carefully to see if the assumption of minority carrier transport is satisfied.
In regions of non-zero field like NP junctions, we can use band diagrams to sketch ℰ, 𝑉, 𝑛, and 𝑝 ; and Poisson’s equation and ‘depletion approximation’ to quantify these values.
This approach also gives us the width of the ‘depletion region’ on both sides of the junction (𝑥𝑛 and 𝑥𝑝), plus the ‘built-in’ voltage 𝑉𝑏𝑖
2/8/18 Bermel ECE 305 S1852